Advertisements
Advertisements
प्रश्न
Solve the following equation by factorization
`(8)/(x + 3) - (3)/(2 - x)` = 2
Advertisements
उत्तर
`(8)/(x + 3) - (3)/(2 - x)` = 2
`(16 - 8x - 3x - 9)/((x + 3)(2 - x)` = 2
⇒ `(-11x + 7)/(2x - x^2 + 6 - 3x)` = 2
⇒ -11x + 7 = 4x - 2x2 + 12 - 6x
⇒ -11x + 7 - 4x + 2x2 - 12 + 6x = 0
⇒ 2x2 - 9x - 5 = 0
⇒ 2x2 - 10x + x - 5 = 0
⇒ 2x(x - 5) + 1(x - 5) = 0
⇒ (x - 5) (2x + 1) = 0
Either x - 5 = 0,
then x = 5
or
2x + 1 = 0,
then 2x = -1
⇒ x = `-(1)/(2)`
Hence x = 5, `-(1)/(2)`.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
4x2 + 5x = 0
Solve the following quadratic equations by factorization:
a2b2x2 + b2x - a2x - 1 = 0
The sum of two number a and b is 15, and the sum of their reciprocals `1/a` and `1/b` is 3/10. Find the numbers a and b.
Solve the following quadratic equation for x:
x2 − 4ax − b2 + 4a2 = 0
Solve the following quadratic equation by factorisation.
m2 - 11 = 0
Solve the following quadratic equation by factorisation method:
`(x + 3)/(x - 2) - (1 - x)/x = (17)/(4)`.
In each of the following, determine whether the given values are solution of the given equation or not:
`a^2x^2 - 3abx + 2b^2 = 0; x = a/b, x = b/a`.
Solve the following equation by factorization
`x^2/(15) - x/(3) - 10` = 0
Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.
Solve the quadratic equation: x2 – 2ax + (a2 – b2) = 0 for x.
